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Algebra Difficulty 4.5 AIME Prove it Saudi Arabia

Prove that for every real number $x$ the following inequality holds:
$x^6 + x^4 - x^3 - x + \frac{3}{4} > 0.$

Solution

The inequality is equivalent to
$x^6 - x^3 + \frac{1}{4} + x^4 - x^2 + \frac{1}{4} + x^2 - x + \frac{1}{4} > 0,$
which in turn is equivalent to
$\left(x^3 - \frac{1}{2}\right)^2 + \left(x^2 - \frac{1}{2}\right)^2 + \left(x - \frac{1}{2}\right)^2 > 0.$

Since $\frac{1}{\sqrt{2}} \neq \frac{1}{2}$ , equality cannot occur.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.